A steel ball is dropped from a height of 50.0m. How far does this ball travel after 3 seconds?
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To find the distance traveled by the steel ball after 3 seconds, we can use the equation of motion:
Distance = initial velocity × time + (1/2) × acceleration × time squared
In this case, the steel ball is dropped, so its initial velocity is 0 m/s. The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the values, we get:
Distance = 0 × 3 + (1/2) × 9.8 × 3^2
= 0 + (1/2) × 9.8 × 9
= 0 + 4.9 × 9
= 44.1 m
Therefore, the steel ball travels a distance of 44.1 meters after 3 seconds.
To find the distance traveled by the ball after 3 seconds, we can use the equation of motion:
\[d = ut + \frac{1}{2} a t^2\]
where:
- \(d\) is the distance traveled by the ball,
- \(u\) is the initial velocity of the ball (which is 0 m/s since the ball is dropped),
- \(t\) is the time (in seconds), and
- \(a\) is the acceleration due to gravity (which is approximately 9.8 m/s²).
In this case, we are given the initial height of the ball (50.0 m), which can be considered as the distance traveled in the downward direction. So, we can rewrite the equation as:
\[d = -\frac{1}{2} g t^2\]
where \(g\) is the acceleration due to gravity.
Substituting the values into the equation:
\[d = -\frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (3 \, \text{s})^2\]
Calculating this expression will give us the distance traveled by the ball after 3 seconds.